Question

solve the separable differential equation
5x - 8y\sqrt{x^2 + 1} \frac{dy}{dx} = 0
subject to the initial condition: y(0) = 3.
y = \square.
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Explanation:

Step1: Separate Variables

Given the differential equation \( 5x - 8y\sqrt{x^2 + 1}\frac{dy}{dx}=0 \), we rearrange to separate variables. Move the terms involving \( y \) and \( dy \) to one side, and terms involving \( x \) and \( dx \) to the other:
\( 8y\sqrt{x^2 + 1}\frac{dy}{dx}=5x \)
\( 8y \, dy=\frac{5x}{\sqrt{x^2 + 1}} \, dx \)

Step2: Integrate Both Sides

Integrate the left - hand side with respect to \( y \) and the right - hand side with respect to \( x \).

• For the left - hand side: \( \int 8y \, dy = 8\int y \, dy \). Using the power rule \( \int y^n dy=\frac{y^{n + 1}}{n+1}+C \) (here \( n = 1 \)), we get \( 8\times\frac{y^{2}}{2}+C_1 = 4y^{2}+C_1 \).
• For the right - hand side: Let \( u=x^{2}+1 \), then \( du = 2x \, dx \) or \( \frac{du}{2}=x \, dx \). So \( \int\frac{5x}{\sqrt{x^2 + 1}} \, dx=5\int\frac{x}{\sqrt{x^2 + 1}} \, dx \). Substituting \( u \) and \( \frac{du}{2} \) into the integral, we have \( 5\int\frac{1}{\sqrt{u}}\times\frac{du}{2}=\frac{5}{2}\int u^{-\frac{1}{2}}du \). Using the power rule \( \int u^n du=\frac{u^{n + 1}}{n + 1}+C \) (here \( n=-\frac{1}{2} \)), we get \( \frac{5}{2}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C_2=5\sqrt{u}+C_2 = 5\sqrt{x^{2}+1}+C_2 \).

Combining the two integrals (we can combine the constants \( C_1 \) and \( C_2 \) into a single constant \( C \)):
\( 4y^{2}=5\sqrt{x^{2}+1}+C \)

Step3: Apply Initial Condition \( y(0) = 3 \)

Substitute \( x = 0 \) and \( y = 3 \) into the equation \( 4y^{2}=5\sqrt{x^{2}+1}+C \):
\( 4\times(3)^{2}=5\sqrt{0^{2}+1}+C \)
\( 4\times9 = 5\times1+C \)
\( 36=5 + C \)
Solve for \( C \): \( C=36 - 5=31 \)

Step4: Solve for \( y \)

Substitute \( C = 31 \) back into the equation \( 4y^{2}=5\sqrt{x^{2}+1}+C \):
\( 4y^{2}=5\sqrt{x^{2}+1}+31 \)
Divide both sides by 4: \( y^{2}=\frac{5\sqrt{x^{2}+1}+31}{4} \)
Since \( y(0) = 3>0 \), we take the positive square root: \( y=\sqrt{\frac{5\sqrt{x^{2}+1}+31}{4}}=\frac{\sqrt{5\sqrt{x^{2}+1}+31}}{2} \)

Answer:

\( y=\frac{\sqrt{5\sqrt{x^{2}+1}+31}}{2} \)